What is the mass of heavy water, d2o(l), produced when 7.60 g of o2(g) reacts with excess d2(g)?

19.026 g

The reaction between D₂ and O₂ is given by the equation;

2D₂(g) + O₂(g) → 2H₂O(l)

We are given;

Mass of oxygen (O₂) that reacted as 7.60 g

We are required to calculate the mass of D₂O produced;

To calculate the number of moles we divide mass by the molar mass.

Moles = mass ÷ Molar mass

Molar mass of O₂ is 16.0 g/mol

Therefore;

Moles = 7.60 g ÷ 16.0 g/mol

         = 0.475 moles

From the equation, 1 mole of oxygen reacts to produce 2 moles of D₂O

Thus, the mole ratio of O₂ : D₂O is 1 : 2

Therefore, moles of D₂O = Moles of O₂ × 2

                                          = 0.475 moles × 2

                                          = 0.95 moles

Mass = Number of moles × Molar mass

Molar mass of heavy water = 20.0276 g/mol

Therefore;

Mass of heavy water = 0.95 moles × 20.0276 g/mol

                                   = 19.026 g

Hence, the mass of heavy water produced is 19.026 g

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.